Law of cosines

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==Proof==
==Proof==
===By the Pythagorean Theorem===
===By the Pythagorean Theorem===
-
An easy way to think of the law of cosines is as an extension of the Pythagorean theorem:
+
An easy way to think of the law of cosines is as an extension of the Pythagorean theorem for a right triangle:
::<math>a^{2} + b^{2} = c^{2} </math>
::<math>a^{2} + b^{2} = c^{2} </math>
-
[[Image:Pythagorean_cosines_proof.jpg]]
+
[[Image:Pythagoras_law_of_cosines.jpg]]
 +
By Pythagorean theorem, we know
 +
::<math> c^{2} = p^{2} + h^{2}</math>
 +
 +
But <math> p</math> is just some portion of side length <math>a</math> which is <math>r</math> less than the length of <math>a</math>. Substituting the difference gives us,
 +
 +
::<math> c^{2} = (a-r)^{2} + h^{2}</math>
 +
 +
By Pythagorean theorem, we also know that
 +
 +
::<math> h^{2} = b^{2} - r^{2}</math>
 +
 +
Substituting the appropriate values gives us,
 +
 +
::<math> c^{2} = (a-r)^{2} + ( b^{2} - r^{2})</math>
 +
 +
Expanding the squared term gives us
 +
 +
::<math> c^{2} = a^{2} -2ar + r^{2} + b^{2} - r^{2}</math>
 +
 +
Simplify for
 +
 +
::<math> c^{2} = a^{2} -2ar + b^{2}</math>
 +
 +
And by the definition of cosine, we know that
 +
 +
::<math>r=b \cos C</math>
 +
 +
Substituting this value in give us
 +
 +
::<math> c^{2} = a^{2} -2ab \cos C + b^{2}</math>
 +
 +
or
 +
 +
::<math> c^{2} = a^{2}+ b^{2} -2ab \cos C </math>
===Using the Distance Formula===
===Using the Distance Formula===
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-
<math> distance = \sqrt {(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}</math>
+
::<math> \text{distance} = \sqrt {(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}</math>
<math>c</math> is the distance from <math> A</math> to <math> B</math>.
<math>c</math> is the distance from <math> A</math> to <math> B</math>.
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Substituting the appropriate points into the distance formula gives us
Substituting the appropriate points into the distance formula gives us
-
<math> c = \sqrt {(a-b \cos C)^{2} + (0-b \sin C)^{2}}</math>
+
::<math> c = \sqrt {(a-b \cos C)^{2} + (0-b \sin C)^{2}}</math>
Squaring the inner terms, we have
Squaring the inner terms, we have
-
<math> c = \sqrt {(a^{2}-2ab \cos C+b^{2} \cos^{2} C) + (b^{2} \sin^{2} C)}</math>
+
::<math> c = \sqrt {(a^{2}-2ab \cos C+b^{2} \cos^{2} C) + (b^{2} \sin^{2} C)}</math>
Since <math> \cos^{2} C + \sin^{2} C = 1</math>,
Since <math> \cos^{2} C + \sin^{2} C = 1</math>,
-
<math> c = \sqrt {a^{2}+b^{2}-2ab \cos C+b^{2}}</math>
+
::<math> c = \sqrt {a^{2}+b^{2}-2ab \cos C+b^{2}}</math>
Square both sides for
Square both sides for
-
<math> c^{2} = a^{2}+b^{2}-2ab \cos C+b^{2}</math>
+
::<math> c^{2} = a^{2}+b^{2}-2ab \cos C+b^{2}</math>
==Example Problem==
==Example Problem==
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-
<math> c^{2} = a^{2} + b^{2} - 2ab \cos C </math>
+
::<math> c^{2} = a^{2} + b^{2} - 2ab \cos C </math>
[[Image:SAS triangle.jpg]]
[[Image:SAS triangle.jpg]]
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-
<math> c^{2} = 6^{2} + (6 \sqrt{2})^{2} -2 (6) (6 \sqrt{2}) \cos 45^\circ </math>
+
:<math> c^{2} = 6^{2} + (6 \sqrt{2})^{2} -2 (6) (6 \sqrt{2}) \cos 45^\circ </math>
Simplify for
Simplify for
-
<math> c^{2} =36 + 36 (2) - 72 \sqrt{2}) \cos 45^\circ </math>
+
::<math> c^{2} =36 + 36 (2) - 72 \sqrt{2}) \cos 45^\circ </math>
Since <math> \cos 45^\circ = \frac{1}{\sqrt{2}}</math>, substitution gives us
Since <math> \cos 45^\circ = \frac{1}{\sqrt{2}}</math>, substitution gives us
-
<math> c^{2} =36 + 36 (2) - 72 \sqrt{2} (\frac{1}{\sqrt{2}}) </math>
+
::<math> c^{2} =36 + 36 (2) - 72 \sqrt{2} (\frac{1}{\sqrt{2}}) </math>
Simplify for
Simplify for
-
<math> c^{2} =36 + 72 - 72 </math>
+
::<math> c^{2} =36 + 72 - 72 </math>
-
<math> c^{2} =36 </math>
+
::<math> c^{2} =36 </math>
Taking the square root of both sides gives us
Taking the square root of both sides gives us
-
<math> c =6 </math>
+
::<math> c =6 </math>
----
----
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-
<math> b^{2} = a^{2} + c^{2} - 2ab \cos B </math>
+
::<math> b^{2} = a^{2} + c^{2} - 2ab \cos B </math>
Substituting in the appropriate side lengths gives us
Substituting in the appropriate side lengths gives us
-
<math> (6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B </math>
+
::<math> (6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B </math>
Simplify for
Simplify for
-
<math> 36 (2) = 36 + 36 - 72 \cos B </math>
+
::<math> 36 (2) = 36 + 36 - 72 \cos B </math>
-
<math>72 = 72 - 72 \cos B </math>
+
::<math>72 = 72 - 72 \cos B </math>
Subtracting <math>72</math> from both sides gives us
Subtracting <math>72</math> from both sides gives us
-
<math>0 = - 72 \cos B </math>
+
::<math>0 = - 72 \cos B </math>
Dividing both sides by <math>-72</math> gives us
Dividing both sides by <math>-72</math> gives us
-
<math>0 = \cos B </math>
+
::<math>0 = \cos B </math>
Using [[Basic Trigonometric Functions#Inverse Trig Functions| inverse trigonometry]], we know that
Using [[Basic Trigonometric Functions#Inverse Trig Functions| inverse trigonometry]], we know that
-
<math>B = 90^\circ </math>
+
::<math>B = 90^\circ </math>
And we can find the last angle measure <math>A</math> by subtracting the other two measures from <math> 180^\circ</math>
And we can find the last angle measure <math>A</math> by subtracting the other two measures from <math> 180^\circ</math>
-
<math> 180^\circ - 90^\circ - 45^\circ = 45^\circ</math>
+
::<math> 180^\circ - 90^\circ - 45^\circ = 45^\circ</math>
-
<math> A=45^\circ</math>
+
::<math> A=45^\circ</math>
[[Image:SAS_solution.jpg]]
[[Image:SAS_solution.jpg]]

Revision as of 11:25, 3 June 2011

Image:inprogress.png
Law of Cosines
[[Image:Image:Law_of_cosines_pictiure.jpg|center|400px]]
Field: Geometry
Created By: [[Author:| ]]
Website: [ ]

Law of Cosines

The law of cosines is a trigonometric extension of the Pythagorean Theorem.


Contents

Basic Description

The law of cosines is a formula that helps in solving triangles when two or three side lengths of a triangle are known. The formula combines the squares of two side lengths of a triangle and some offset, classified by the cosine of a particular angle, to calculate the square of the third side. For this reason, the law of cosines is often thought of as the generalization of the Pythagorean theorem, which only applies to right triangles. The law of cosines adds an extra term to the Pythagorean theorem so that a third side length of a triangle can be determined when there is no right angle.
c^{2} = a^{2} + b^{2} - 2ab \cos C

When to use it: SAS, SSS.

A More Mathematical Explanation

Proof

By the Pythagorean Theorem

An easy way to think of the law of cosines is as an extens [...]

Proof

By the Pythagorean Theorem

An easy way to think of the law of cosines is as an extension of the Pythagorean theorem for a right triangle:

a^{2} + b^{2} = c^{2}

Image:Pythagoras_law_of_cosines.jpg

By Pythagorean theorem, we know

c^{2} = p^{2} + h^{2}

But p is just some portion of side length a which is r less than the length of a. Substituting the difference gives us,

c^{2} = (a-r)^{2} + h^{2}

By Pythagorean theorem, we also know that

h^{2} = b^{2} - r^{2}

Substituting the appropriate values gives us,

c^{2} = (a-r)^{2} + ( b^{2} - r^{2})

Expanding the squared term gives us

c^{2} = a^{2} -2ar + r^{2} + b^{2} - r^{2}

Simplify for

c^{2} = a^{2} -2ar + b^{2}

And by the definition of cosine, we know that

r=b \cos C

Substituting this value in give us

c^{2} = a^{2} -2ab \cos C + b^{2}

or

c^{2} = a^{2}+ b^{2} -2ab \cos C

Using the Distance Formula

Let \vartriangle ABC be oriented so that C is at the origin, and B is at the point (a,0).

Image:Law_of_cosines_proof.jpg


\text{distance} = \sqrt {(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}

c is the distance from A to B.

Substituting the appropriate points into the distance formula gives us

c = \sqrt {(a-b \cos C)^{2} + (0-b \sin C)^{2}}

Squaring the inner terms, we have

c = \sqrt {(a^{2}-2ab \cos C+b^{2} \cos^{2} C) + (b^{2} \sin^{2} C)}

Since \cos^{2} C + \sin^{2} C = 1,

c = \sqrt {a^{2}+b^{2}-2ab \cos C+b^{2}}

Square both sides for

c^{2} = a^{2}+b^{2}-2ab \cos C+b^{2}

Example Problem

Complete the triangle using the law of cosines.


c^{2} = a^{2} + b^{2} - 2ab \cos C

Image:SAS triangle.jpg

Solution

To find the side length c,


c^{2} = 6^{2} + (6 \sqrt{2})^{2} -2 (6) (6 \sqrt{2}) \cos 45^\circ

Simplify for

c^{2} =36 + 36 (2) - 72 \sqrt{2}) \cos 45^\circ

Since \cos 45^\circ = \frac{1}{\sqrt{2}}, substitution gives us

c^{2} =36 + 36 (2) - 72 \sqrt{2} (\frac{1}{\sqrt{2}})

Simplify for

c^{2} =36 + 72 - 72


c^{2} =36

Taking the square root of both sides gives us

c =6

Now we can orient the triangle differently to get get a new version of the law of cosines so we can find angle measure B,


b^{2} = a^{2} + c^{2} - 2ab \cos B

Substituting in the appropriate side lengths gives us

(6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B

Simplify for

36 (2) = 36 + 36 - 72 \cos B
72 = 72 - 72 \cos B

Subtracting 72 from both sides gives us

0 = - 72 \cos B

Dividing both sides by -72 gives us

0 = \cos B

Using inverse trigonometry, we know that

B = 90^\circ

And we can find the last angle measure A by subtracting the other two measures from 180^\circ

180^\circ - 90^\circ - 45^\circ = 45^\circ
A=45^\circ
Image:SAS_solution.jpg




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